Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Applications of Derivatives Important Questions which are most likely to be asked in the exam.

## Intermediate 1st Year Maths 1B Applications of Derivatives Important Questions

Question 1.

Find dy and ∆y of y = f(x) = x^{2} + x at x = 10 when ∆x = 0.1.

Solution:

As change in y – f(x) is given by ∆y = f(x + ∆x) – f(x), this change at x = 10 with ∆x = 0.1 is

∆y = f(10.1) – f(10)

= {(10.1)^{2} + 10.1} – {10^{2} + 10}

= 2.11.

Since dy = f'(x) ∆x, dy at x = 10 with ∆x = 0.1 is dy = {(2)(10) + 1} 0.1 = 2.1

(since \(\frac{\mathrm{d} y}{\mathrm{dx}}\) = 2x + 1).

Question 2.

Find ∆y and dy for the function y = cos (x) at x = 60° with ∆x = 1° .

Solution:

For the given problem ∆y and dy at x = 60° with ∆x = 1° are

∆y = cos (60° + 1°) – cos (60°) …………. (1)

and dy = -sin(60°) (1°) ………….. (2)

Cos (60°)= 0.5,

Cos (61°)= 0.4848,

Sin (60°) = 0.8660,

1° = 0.0174 radians

∴ ∆y = -0.0152 and

dy = -0.0150.

Question 3.

The side of a square is increased from 3 cm to 3.01 cm. Find the approximate increase in the area of the square.

Solution:

Let x be the side of a square and A be its area.

Then A = x^{2}. …… (1)

Clearly A is a function of x. As the side is increased from 3 cm to 3.01 cm we can take x – 3 and ∆x = 0.01 to compute the approximate increase in the area of square. The approximate value of change in area is

∆A ≈ \(\frac{\mathrm{dA}}{\mathrm{dx}}\) ∆x

In view of equation (1), the equation (2) becomes

∆A ≈ 2x∆x

Hence the approximate increase in the area when the side is increased from 3 to 3.01 is

∆A ≈ 2(3)(0.01) = 0.06

Question 4.

If the radius of a sphere is increased from 7 cm to 7.02 cm then find the approximate increase in the volume of the sphere.

Solution:

Let r be the radius of a sphere and V be its volume. Then

V = \(\frac{4 \pi \pi^{2}}{3}\) …………….. (1)

Here V is a function of r. As the radius is increased from 7 cm to 7.02, we can take r = 7 cm and ∆r = 0.02 cm. Now we have to find the approximate increase in the volume of the sphere.

∴ ∆V ≈ \(\frac{\mathrm{dV}}{\mathrm{dr}}\) ∆r = 4πr^{2} ∆r.

Thus, the approximate increase in the volume of the sphere is \(\frac{4(22)(7)(7)(0.02)}{7}\) = 12.32 cm^{3}.

Question 5.

If y = f(x) = k x^{n} then show that the approximate relative error (or increase) in y is n times the relative error (or increase) in x where n and k are constants.

Solution:

The approximate relative error (or increase) in y by the equation (2) of if a number A is very close to a number B but it is not equal to B then A is called an approximate value of B is (\(\frac{f^{\prime}(x)}{f(x)}\)) ∆x = \(\frac{k n x^{n-1}}{k x^{n}}\) ∆x = n(\(\frac{\Delta x}{x}\) = n)

Hence the approximate relative error in y = kx^{n} is n times the relative error in x.

Question 6.

If the increase in the side of a square is 2% then find the approximate percentage of increase in its area.

Solution:

Let x be the side of a square and A be its area.

Then A = x^{2}.

Approximate percentage error in area A

= (\(\frac{\frac{\mathrm{dA}}{\mathrm{dx}}}{\mathrm{A}}\)) × 100 × ∆x(by (3) of if a number

A is very close to a number B but it is not equal to B then A is called an approximate value of B with f = A)

= \(\frac{100(2 x) \Delta x}{x^{2}}\) = \(\frac{200 \Delta x}{x}\) = 2(2) = 4

(∵ \(\frac{\Delta x}{x}\) × 100 = 2

Question 7.

If an error of 0.01 cm is made in measuring the perimeter of a circle and the perimeter is measured as 44 cm then find the approximate error and relative error in its area.

Solution:

Let r, p and A be the radius, perimeter and area of the circle respectively. Given that p = 44 cm and ∆p = 0.01. We have to find approximation of ∆A and \(\frac{\Delta \mathrm{A}}{\mathrm{A}}\). Note that A = πr^{2} which is a function of r. As p and ∆p are given we have to transform A = πr^{2} into the form A = f(p). This can be achieved by using the relation, perimeter 2πr = p.

∴ A = π(\(\frac{p}{2 \pi}\))^{2} = \(\frac{p^{2}}{4 \pi}\)

Hence the approximate error in

A = \(\frac{d A}{d p}\)∆p = \(\frac{2 p}{4 \pi}\)∆p = \(\frac{P}{2 \pi}\)∆p

The approximate error in A when p = 44 and ∆p = 0.01 = \(\) (0.01) = 0.07

The approximate relative error

= 0.0004545.

Question 8.

Find the approximate value of \(\sqrt[3]{999}\)

Solution:

This problem can be answered by

f(x + ∆x) ≈ f(x) + f'(x) ∆x …………. (1)

with x = 1000 and ∆x = -1. The reason for taking x = 1000 is to make the calculation of f(x) simpler when f(x) = \(\sqrt[3]{x}\), Suppose

y = f(x) = \(\sqrt[3]{x}\)

The equation (1) becomes

f(x + ∆x) ≈ f(x) = \(\frac{1}{3 x^{\frac{2}{3}}}\) ∆x

Hence f(1000 – 1)

≈ f(1000) + \(\frac{1}{3(1000)^{2 / 3}}\) (-1) = 9.9967.

Question 9.

Find the slope of the tangent to the following curves at the points as indicated.

i) y = 5x^{2} at (-1, 5)

ii) y = \(\frac{1}{x – 1}\) (x ≠ 1) at [3, \(\frac{1}{2}\)]

iii) x = a secθ, y = a tanθ at θ = \(\frac{\pi}{6}\)

iv) (\(\frac{x}{a}\))^{n} + (\(\frac{x}{b}\))^{n} = 2 at (a, b)

Solution:

i) y = 5x^{2}, then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 10x

Slope of the tangent at the given points

(\(\frac{\mathrm{dy}}{\mathrm{dx}}\))_{(-1.5)} -10(-1) = -10

ii) y = \(\frac{1}{x – 1}\)

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-1}{(x-1)^{2}}\)

Slope of the tangent at the given point is

Question 10.

Find the equations of the tangent and the normal to the curve y = 5x^{4} at the point (1, 5)

Solution:

y = 5x^{4} implies that \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 20x^{3}

Slope of the tangent to the curve at (1, 5) is

(\(\frac{\mathrm{dy}}{\mathrm{dx}}\))_{(1, 5)} = 20(1)^{5} = 20

Equation of the normal of (1, 5) is

y – 5 = 20 (x – 1) = 20x – 20

y = 20x – 15

Slope of the normal to the curve at (1, 5) is

–\(\frac{1}{m}\) = –\(\frac{1}{20}\)

Equation of the normal of (1, 5) is

y – 5 = –\(\frac{1}{20}\) (x – 1)

20 y – 100 = -x + 1

x + 20y = 101

(or) 20y = 101 – x

Question 11.

Find the equations of the tangent and the normal to the curve y^{4} = ax^{3}. at (a, a).

Solution:

Given curve is y^{4} = ax^{3}

Differentiating w.r.to. x

4y^{3} . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3ax^{2}

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{3 a x^{2}}{4 y^{3}}\)

Slope of the tangent at (a, a) = \(\frac{3 a \cdot a^{2}}{4 a^{3}}\) = \(\frac{3}{4}\)

Slope of the normal at (a, a) = –\(\frac{1}{m}\) = –\(\frac{4}{3}\)

Equation of the tangent at (a, a) is

y – a = \(\frac{3}{4}\)(x – a)

4y – 4a = 3x – 3a

4y = 3x + a

Equation of the normal at (a, a) is 4

y – a = – \(\frac{4}{3}\)(x – a)

3y – 3a = -4x + 4a

3y + 4x = 7a

Question 12.

Find the equations of the tangent to the curve y = 3x^{2} – x^{3}, where it meets the X-axis.

Solution:

Equation of the curve is y = 3x^{2} – x^{3}

Equation of X – axis is y = 0

For points is intersection of the curve and X-axis

3x^{2} – x^{3} = 0 ⇒ x^{2} (3 – x) = 0

x = 0, x = 3

The curve crosses X-axis 0(0, 0) and A(3, 0)

y = -3x^{2} – x^{3}

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 6x – 3x^{2}

At O(0, 0), slope of the tangent = (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))_{(0, 0)} = 0

Equation of the tangent at (0, 0) is y – 0, 0(x – 0) i.e., y = 0

i.e., x-axis is the tangent to the curve at O(0, 0)

At A(3, 0), slope of the tangent = (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))_{(3, 0)}

= 6.3 – 3.3^{2}

= 18 – 27

= -9

Equation of the tangent at A(3, 0) is

y – 0 = -9(x – 3) = -9x + 27

(or) 9x + y = 27

Question 13.

Find the points at which the curve t y = sin x has horizontal tangents.

Solution:

y = sin x

A tangent is horizontal if and anal its slope is

cos x = 0 ⇒ x = (2n + 1)\(\frac{\pi}{2}\), n ∈ Z

Hence the given curve has horizontal tangents at points (x_{0}, y_{0})

⇔ x_{0} = (2n + 1) . \(\frac{\pi}{2}\) and

y_{0} = (-1)^{n} for same n ∈ Z

Question 14.

Verify whether the curve y = f(x) = x^{1/3} has a vertical tangent at the point with x = 0.

Solution:

The function has a verified tangent at the point whose x co-ordinate is 0.

Question 15.

Find whether the curve y = f(x)= x^{2/3} has a vertical tangent at x = 0.

Solution:

Thus left handed be normal \(\frac{1}{h^{1 / 3}}\) as h → 0 is -∞

While the right handed limit is ∞.

Hence does not exist. The vertical tangent does not exist.

At the point x = 0.

Question 16.

Show that the tangent at any point 0 on the curve x = c sec θ, y = c tan θ is y sin θ = x – c cos θ.

Solution:

x = c sec θ, y = c tan θ

Question 17.

Show that the area of the triable formed by the tangent at any point on the curve xy = c (c ≠ 0) with the coordinate axis is constant.

Solution:

Observe that c ≠ 0

If c = 0 the equation xy = 0 represent the coordinate circle which is against the definite.

Let P (x_{1}, y_{1}) be a point on the curve xy = c

y = \(\frac{c}{x}\) = 1, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = – \(\frac{c}{x^{2}}\)

Equation of the tangent at (x_{1}, y_{1}) is

y – y_{1} = – \(\frac{c}{x^{2}}\) (x – x_{1})

x^{2}y – x_{1}^{2} = -cx + cx_{1}

cx + x_{1}^{2} . y = x_{1}^{2} + cx_{1}

= cx_{1} + cx_{1} (x_{1}y_{1} = c)

= 2cx_{1}

\(\frac{c x}{2 c x_{1}}\) + \(\frac{x_{1}^{2} y}{2 c x_{1}}\) = 1

\(\frac{x}{2 x_{1}}\) + \(\frac{y}{\left(\frac{2 c}{x_{1}}\right)}\) = 1

Area of the triangle formed with co-ordinate axes

= \(\frac{1}{2}\) |OA . OB|

= \(\frac{1}{2}\) (2x_{1}) (\(\frac{2 c}{x_{1}}\)) = 2c = constant

Question 18.

Show that the equation of the tangent to the curve (\(\frac{x}{a}\))^{n} + (\(\frac{y}{b}\))^{n} = 2 (a ≠ 0, b ≠ 0) at the point (a, b) is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 2

Solution:

Equation of the tangent to the curve at the point (a, b) is

y – b = \(\frac{-b}{a}\) (x – a)

\(\frac{y}{b}\) – 1 = – \(\frac{x}{a}\) + 1

\(\frac{x}{a}\) + \(\frac{y}{b}\) = 2

Question 19.

Show that the length of the sub normal at point on the curve y^{2} = 4ax is a constant.

Solution:

Equation of the curve is y^{2} = 4ax

Differentiating w.r.to x

2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 4a

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{4a}{2y}\) = \(\frac{2a}{y}\)

Length of the sub-normal \(\left|\frac{y d y}{d x}\right|=\left|y \cdot \frac{2 a}{y}\right|\)

= 2a = constant

Question 20.

Show that the length of the Sub tangent at any point on the curve y = a^{x} (a > 0) is a constant.

Solution:

Equation of the curve is y = a^{x}

Differentiating w.r.to x

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = a^{x} log a = y. log a

Length of the sub-tangent

= \(\left|\frac{y}{\left(\frac{d y}{d x}\right)}\right|\) = \(\left|\frac{y}{\left(\frac{d y}{d x}\right)}\right|\) = \(\frac{1}{\log a}\) constant

Question 21.

Show that the square of the length of subtangent at any point on the curve by^{2} = (x + a)^{3} (b ≠ 0) varies with the length of the subnormal at that point.

Solution:

Differentiating by^{2} = (x + a)^{3}

w.r.t x, we get

2by y’ = 3(x + a)^{2}

∴ The length of the subnormal at any point (x, y) on the curve

= |y y’| = |\(\frac{3}{2 b}\)(x + a)^{2}| ………………… (1)

The square of the length of subtangent

∴ (length of the subnormal)^{2} ∝ (length of subnormal).

Question 22.

Find the value of k, so that the length of the subnormal at any point on the curve y = a^{1 – k}x^{k} is a constant.

Solution:

Differentiating y = a^{1 – k}x^{k} with respect to x,

we get y’ = ka^{1 – k}x^{k – 1}

Length of subnormal at any point P(x, y) on the curve

= |y y’| = |yka^{1 – k}x^{k – 1}|

= |k a^{1 – k}x^{k}a^{1 – k}x^{k – 1}|

= |ka^{2-2k}x^{2k-1}|

In order to make these values a constant, we should have 2k – 1 = 0 i.e., k = \(\frac{1}{2}\).

Question 23.

Find the angle between the curves xy = 2 and x^{2} + 4y = 0.

Solution:

First we find the points of intersection of xy = 2 and x^{2} + 4y = 0

y = \(\frac{-x^{2}}{4}\)

But xy = 2 ⇒ x(\(\frac{-x^{2}}{4}\)) = 2 ⇒ x^{3} = -8

x = -2

y = \(\frac{-x^{2}}{4}\) = – \(\frac{4}{4}\) = -1

Point of intersection is P(-2, -1)

= \(\left|\frac{-\frac{3}{2}}{\frac{1}{2}}\right|\) = 3

Φ = tan^{-1}(3)

Question 24.

Find the angle between the curve 2y = e^{\(\frac{-x}{2}\)} and Y-axis.

Solution:

Equation of Y-axis is x = 0.

The point of intersection of the curve

2y = e^{\(\frac{-x}{2}\)} and x = 0 is P(0, \(\frac{1}{2}\))

Let ψ be the angle between the given curves

2y = e^{\(\frac{-x}{2}\)} at P with X – axis is given by

tan ψ = \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\left(0, \frac{1}{2}\right)}=\left.\frac{-1}{4} \mathrm{e}^{\frac{-x}{2}}\right|_{\left(0, \frac{1}{2}\right)}=\frac{-1}{4}\)

Further, if Φ is the angle between the Y – axis and 2y = e\(\frac{-x}{2}\), then we have

tan Φ = |tan (\(\frac{\pi}{2}\) – ψ)| = |cot ψ| = 4

∴ The angle between the curve and the Y-axis is tan^{-1} 4.

Question 25.

Show that the condition of the orthogonality of the curves ax^{2} + by^{2} = 1 and a_{1}x^{2} + b_{1}y^{2} = 1 is \(\frac{1}{a}\) – \(\frac{1}{b}\) = \(\frac{1}{a_{1}}\) – \(\frac{1}{b_{1}}\).

Solution:

Let the curves ax^{2} + by^{2} = 1 and a_{1}x^{2} + b_{1}y^{2} = 1 intersect at p(x_{1}, y_{1}) so that

ax_{1}^{2} + by_{1}^{2} = 1 and a_{1}x_{1}^{2} + b_{1}y_{1}^{2} = 1, from which we get,

\(\frac{x_{1}^{2}}{b_{1}-b}\) = \(\frac{y_{1}^{2}}{a_{1}-a}\) = \(\frac{1}{a b_{1}-a_{1} b}\) …………. (1)

Differentiating ax^{2} + by^{2} = 1 with respect to x,

we get \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-a x}{b y}\)

Hence, if mt is the slope of the tangent at P(x_{1}, y_{1}) to the curve

ax^{2} + by^{2} = 1, m_{1} = \(\frac{-a x_{1}}{b y_{1}}\)

Similarly, the slope (m_{2}) of the tangent at P to

a_{1}x^{2} + b_{1}y^{2} = 1 is given by m_{2} = \(\frac{-a_{1} x_{1}}{b_{1} y_{1}}\)

Since the curves cut orthogonally we have m_{1}m_{2} = -1,

i.e., \(\frac{\mathrm{a} a_{1} x_{1}^{2}}{\mathrm{~b} \mathrm{~b}_{1} y_{1}^{2}}\) = -1 or \(\frac{x_{1}^{2}}{y_{1}^{2}}=\frac{-b_{1}}{a a_{1}}\) ………………….. (2)

Now from (1) and (2), the condition for the orthogonality of the given curves is

\(\frac{b_{1}-b}{a-a_{1}}\) = \(\frac{b b_{1}}{a a_{1}}\)

or (b – a) a_{1}b_{1} = (b_{1} – a_{1}) ab

or \(\frac{1}{a}\) – \(\frac{1}{b}\) = \(\frac{1}{a_{1}}\) – \(\frac{1}{b_{1}}\)

Question 26.

Show that the curves y^{2} = 4(x + 1)and y^{2} = 36 (9 – x) intersect orthogonally. [Mar 11, May 06, 05]

Solution:

Solving y^{2} = 4(x + 1) and y^{2} = 36 (9 – x) for the points of intersection, we get

4(x + 1) = 36 (9 – x) 10x = 80 or x = 8

i.e., y^{2} = 4(x + 1) ⇒ y^{2} = 4(9) = 36 ⇒ y = ± 6

The points of intersection of the two curves are P(8, 6), Q(8, -6)

y^{2} = 4(x + 1) ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2}{y}\)

y^{2} = 36 (9 – x) ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-18}{y}\)

Slope of the tangent to the curve

y^{2} = 4(x + 1) at P is

m_{1} = \(\frac{2}{6}\) = \(\frac{1}{3}\)

Slope of the tangent to the curve

y^{2} = 36 (9 – x) at P is

m_{2} = \(\frac{-18}{6}\) = -3

m_{1}m_{2} = \(\frac{1}{3}\) × -3 = -1

⇒ the curves intersect orthogonally at P.

We can prove, similarly,that the curves intersect orthogonally at Q also.

Question 27.

Find the average rate of change of s = f (t) = 2t^{2} + 3 between t = 2 and t = 4.

Solution:

The average rate of change of s between t = 2 and t = 4 is \(\frac{f(4)-f(2)}{4-2}\) = \(\frac{35-11}{4-2}\) = 12.

Question 28.

Find the rate of change of area of a circle w.r.t. radius when r = 5 cm.

Solution:

Let A be the area of the circle with radius r.

Then A = πr^{2}. Now, the rate of change of area A w.r.t. r is given by \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 2πr. When r = 5 cm.

\(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 10 π.

Thus, the area of the circle is changing at the rate of 10 π cm^{2}/cm.

Question 29.

The volume of a cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of the edge is 10 centimeters ?

Solution:

Let x be the length of the edge of the cube, V be its volume and S be its surface area. Then, V = x^{3} and S = 6x^{2}. Given that rate of change of volume is 9 cm^{3}/sec.

Therefore, \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 9 cm^{3}/sec.

Now differentiating V w.r.t. t, we get,

Question 30.

A particle is moving in a straight line so that after t seconds its distance is s (in cms) from a fixed point on the line is given by s = f(t) = 8t + t^{3}. Find (i) the velocity at time t = 2 sec (ii) the initial velocity (iii) acceleration at t = 2 sec.

Solution:

The distance s and time t are connected by the relation.

s = f(t) = 8t + t^{3} …………. (1)

∴ velocity (ν) = 8 + 3t^{2} ……………. (2)

and the acceleration is given by

a = \(\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}\) = 6t

i) The velocity at

t = 2 is 8 + 3 (4) = 20cm/sec.

ii) The initial velocity (t = 0) is 8 cm/sec.

iii) The acceleration at t = 2 is 6(2) = 12 cm/sec^{2}

Question 31.

A container in the shape of an inverted cone has height 12 cm and radius 6 cm at the top. If it is filled with water at the rate of 12 cm^{3}/sec., what is the rate of change in the height of water level when the tank is filled 8 cm ?

Solution:

Let OC be height to water level at t sec. The triangles OAB ad OCD are similar triangles.

Let OC = h and CD = r. Given that AB = 6 cm, OA = 12 cm.

\(\frac{r}{6}\) = \(\frac{h}{12}\)

Hence, the rate of change of water level is \(\frac{3}{4 \pi}\) cm/sec when the water level of the tank is 8 cm.

Question 32.

A particle is moving along a line according to s = f (t) = 4t^{3} – 3t^{2} + 5t -1 where s is measured in meters and t is measured in seconds. Find the velocity and acceleration at time t. At what time ‘ the acceleration is zero.

Solution:

Since f(t) = 4t^{3} – 3t^{2} + 5t – 1, the velocity at time t is

v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 12t^{2} – 6t + 5

and the acceleration at time t is a = \(\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}\) = 24t – 6.

The acceleration is 0 if 24t – 6 = 0

i.e., t = \(\frac{1}{4}\)

The acceleration of the particle is zero at

t = \(\frac{1}{4}\) sec.

Question 33.

The quantity (in mg) of a drug in the blood at time t (sec) is given by q = 3(0.4)^{t}. Find the instantaneous rate of change at t = 2 sec.

Solution:

Given that q = 3(0.4)^{t}

∴ \(\frac{\mathrm{dQ}}{\mathrm{dx}}\) = 3(0.4)^{t} log_{e}(0.4) is the instaneous rate of change in q. Hence the instaneous rate of q at time t = 2 sec. is given by

\(\left(\frac{\mathrm{dQ}}{\mathrm{dx}}\right)_{t=2}\) = 3(0.4)^{2} log_{e} (0.4).

Question 34.

Let a kind of bacteria grow by t^{3} (t in sec). At what time the rate of growth of the bacteria is 300 bacteria per sec ?

Solution:

Let g be the amount of growth of bacteria at t sec. Then

g(t) = t^{3} ………… (1)

The growth rate at time t is given by

g'(t) = 3t^{2}

300 = 3t^{2} (given that growth rate is 300)

t = 10 sec.

∴ After t = 10 sec, the growth rate of bacteria should be 300 bacteria/sec.

Question 35.

The total cost C(x) in rupees associated with production of x units of an item is given by C(x) = 0.005 x^{3} – 0.02x^{2} + 30x + 500. Find the marginal cost when 3 units are produced (marginal cost is the rate of change of total cost).

Solution:

Let M represent the marginal cost. Then

M = \(\frac{\mathrm{dC}}{\mathrm{dx}}\)

Hence,

M = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(0.005x^{3} – 0.02x^{2} + 30x + 500) dx

= 0.005(3x^{2}) – 0.02(2x) + 30

∴ The Marginal cost at x = 3 is

(M)_{x = 3} = 0.005 (27) – 0.02 (6) + 30 = 30.015.

Hence the required marginal cost is Rs. 30.02 to produce 3 units.

Question 36.

The total revenue in rupees received from the sale of x units of a product is given by R(x) = 3x^{2} + 36x + 5. Find the marginal revenue when x = 5 (marginal revenue is the rate of change of total revenue).

Solution:

Let m denote the marginal revenue. Then

m = \(\frac{\mathrm{dR}}{\mathrm{dx}}\) (since the total revenue is R(x))

Given that R(x) = 3x^{2} + 36x + 5

∴ m = 6x + 36

The marginal revenue at x = 5 is

[m = \(\frac{\mathrm{dR}}{\mathrm{dx}}\)]_{x = 5} = 30 + 36 = 66

Hence the required marginal revenue is Rs. 66.

Question 37.

Verify Rolle’s theorem for the function y = f(x) = x^{2} + 4 in [-3, 3].

Solution:

Here f(x) = x^{2} = 4. f is continuous on [-3, 3] as x^{2} + 4 is a polynomial which is continuous on any closed interval. Further f(3) = f(-3) = 13 and f is differentiable on [-3, 3].

∴ By Rolle’s theorem ∃ c ∈ (-3, 3) such that f'(c) = 0

The point c = 0 ∈ (-3, 3). Thus Rolle’s theorem is verified.

Question 38.

Verify Rolle’s theorem for the function f(x) = x(x + 3)e^{-x/2} in [-3, 0].

Solution:

Here f(-3) = 0 and f(0) = 0.

We have

f'(x) = \(\frac{\left(-x^{2}+x+6\right)}{2} e^{\frac{-x}{2}}\)

f'(x) = 0 ⇔ -x^{2} + x + 6 = 0 ⇔ x = -2 or 3. Of these two values -2 is in the open interval (-3, 0) which satisfies the conclusion of Rolle’s theorem.

Question 39.

Let f(x) = (x -1) (x – 2) (x – 3). Prove that there is more than one ‘c’ in (1, 3) such that f'(c) = 0. [Mar 13]

Solution:

Observe that f is continuous on (1, 3) differentiable in (1, 3) and f(1) = f(3) = 0.

f'(x) = (x – 1) (x – 2) + (x – 1) (x – 3) + (x – 2) (x – 3)

= 3x^{2} – 12x + 11

\(\frac{12 \pm \sqrt{144-132}}{6}\) = 2 ± \(\frac{1}{\sqrt{3}}\)

Both these roots lie in the open interval (1, 3) and are such that the derivative vanishes at these points.

Question 40.

On the curve y = x^{2}, find a point at which the tangent is parallel to the chord joining (0, 0) and (1, 1).

Solution:

The slope of the chord is \(\frac{1-0}{1-0}\) = 1.

The derivative is \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2x.

We want x such that 2x = 1

i.e., x = \(\frac{1}{2}\)

we not that \(\frac{1}{2}\) is in the open interval (0, 1), as required in the Lagrange’s mean value theorem.

The corresponding point on the curve is (\(\frac{1}{2}\), \(\frac{1}{4}\)).

Question 41.

Show that f(x) = 8x + 2 is a strictly increasing function on R without using the graph of y = f(x).

Solution:

Let x_{1}, x_{2} ∈ R with x_{1} < x_{2}. Then 8x_{1} < 8x_{2}. Adding 2 to both sides of this inequality, we have 8x_{1} + 2 < 8x_{2} + 2. i.e., f(x_{1}) < f(x_{2}).

Thus

x_{1} < x_{2} ⇒ f(x_{1}) < f(x_{2}) ∀ x_{1}, x_{2} ∈ R.

Therefore, the given function f is strictly increasing on R.

Question 42.

Show that f (x) = e^{x} is strictly increasing on R (without graph).

Solution:

Let x_{1}, x_{2} ∈ R such that x_{1} < x_{2}. we know that if a > b then e^{a} > e^{b}

∴ x_{1} < x_{2} ⇒ e^{x1} < e^{x2}

i.e., f(x_{1}) < f(x_{2}).

Hence the given function f is a strictly increasing function.

Question 43.

Show that f(x) = – x + 2 is strictly decreasing on R.

Solution:

Let x_{1}, x_{2} ∈ R x_{1} < x_{2}.

Then x_{1} < x_{2}

⇒ -x_{1} > -x_{2}

⇒ -x_{1} + 2 > -x_{2} + 2

⇒ f(x_{1}) > f(x_{2}).

Therefore the given function f is strictly decreasing on R.

Question 44.

Find the intervals on which

f(x) = x^{2} – 3x + 8 is increasing or decreasing ?

Solution:

Given fucntion is f(x) = x^{2} – 3x + 8.

Differentiating it w.r.t. x, we get f'(x) = 2x – 3

f(x) = 0 for x = 3/2.

since f'(x) < 0 in (-∞, 3/2) the function f(x) is strictly decreasing on (-∞, \(\frac{3}{2}\)) Further since f'(x) > 0 in (\(\frac{3}{2}\), -∞), the function f(x) is a strictly increasing function (\(\frac{3}{2}\), -∞).

Question 45.

Show that f(x) = |x| is strictly decreasing on (-∞, 0) and strictly increasing on (0, ∞).

Solution:

The given function is f(x) = |x| i.e.,

Thus f'(c) = 1 if c > 0, f'(c) = -1 if c < 0. Since f'(c) > 0 on (0, ∞), the function f(x) is strictly increasing on (0, ∞). Since f'(c) < 0 on (-∞, 0), the function f(x) is strictly decreasing on (-∞, 0).

Question 46.

Find the intervals on which the function f(x) = x^{3} + 5x^{2} – 8x + 1 is a strictly increasing function.

Solution:

Given that f(x) = x^{3} + 5x^{2} – 8x + 1.

∴ f'(x) = 3x^{2} + 10x – 8 = (3x – 2) (x + 4)

= 3(x – \(\frac{2}{3}\)) (x – (-4)).

f'(x) is negative in (-4, \(\frac{2}{3}\)) and positive in (-∞, -4) ∪ (\(\frac{2}{3}\), ∞) .

∴ The function is strictly deceasing in (-4, \(\frac{2}{3}\)) and is strictly decreasing in (-∞, -4)and (\(\frac{2}{3}\), ∞)

Question 47.

Find the Intervals on which f(x) = x^{x} (x > 0) is increasing and decreasing.

Solution:

Taking logarithms on both sides of f(x) = x^{x}

we get

log (f(x)) = x log x. Differenetiating it w.r.t. x

we have \(\frac{1}{f(x)}\) f'(x)= 1 + log x

∴ f'(x) = x^{x}( 1 + log x)

f'(x) = 0 ⇒ x^{x}(1 + log x) = 0 ……………. (1)

⇒ 1 + log x = 0

⇒ x = 1/e

Suppose x < 1/e log x < log (1/e) (since the base e > 1). i.e., log x < -1

1 + log x < 0 ⇒ x^{x} (1 + log x) < 0. i.e., f'(x) < 0 Now suppose, x > 1/e. Then log x > log (1/e)

i.e., log x > – 1.

⇒ 1 + log x > 0

⇒ x^{x} (1 + log x) > 0

⇒ f'(x) > 0

Hence, f is strictly decreasing on (o, 1/e) and it is strictly increasin on (1/e, ∞).

Question 48.

Determine the intervals in which f(x) = \(\frac{2}{(x-1)}\) + 18x ∀ x ∈ R \ {0} is strictly increasing and decreasing.

Solution:

Given that f(x) = \(\frac{2}{(x-1)}\) + 18x. Differenetiating

it w.r.t. x, we get

f'(x) = \(\frac{-1}{(x-1)^{2}}\) . 2 + 18 and f'(x) = 0

⇒ \(\frac{2}{(x-1)^{2}}\) = 18 ⇒ (x – 1)^{2} = 1/9.

∴ f'(x) = 0 if x – 1 = 1/3 or x- 1 = -(1/3).

i.e., x = 4/3 or x – 2/3.

The derivative of f(x) can be expressed as

f'(x) = \(\frac{18}{(x-1)^{2}}\) . (x – /3) (x – 4/3)

∴ The given function f(x) is strictly increasing on (-∞, \(\frac{2}{3}\)) and (\(\frac{4}{3}\), ∞) and it is strictly decreasing on (\(\frac{2}{3}\), \(\frac{4}{3}\)).

Question 49.

Let f(x) = sin x – cos x be defined on [0, 2π]. Determine the intervals in which f(x) is strictly decreasing and strictly increasing.

Solution:

Given that f(x) = sin x – cos x.

∴ f'(x) = cos x + sin x

∴ f'(x) = \(\sqrt{2}\) . sin(x + π/4)

Let 0 < x < 3π/4. Then π/4 < x + π/4 < π. ∴ sin (x + π/4) > 0 i.e., f'(x) > 0.

Similarly it can be shown that f'(x) < 0 in (3π/4 . 7π/4) and f'(x) > 0 in (7π/4, 2π).

Thus the function f(x) strictly increasing in (0, \(\frac{3\pi}{24}\) and (\(\frac{7\pi}{4}\), 2π) it is strictly decreasing in (\(\frac{3\pi}{4}\), \(\frac{7\pi}{4}\)).

Question 50.

If 0 ≤ x ≤ \(\frac{\pi}{2}\) then show that x ≥ sinx.

Solution:

Let f(x) = x – sin x. Then f'(x) = 1 – cos x ≥ 0 ∀ x

∴ f is an increasing function for all x.

Now, f(0) = 0. Hence f(x) ≥ f(0) for all x ∈ (0, \(\frac{\pi}{2}\)). Therefore, x ≥ in x.

Question 51.

Let f : R → R be defined by f(x) = 4x^{2} – 4x + 11. Find the global minimum value and a point of global minimum.

Solution:

We have to look for a value c e R(domain) such that

f(x) ≥ f(c) ∀ x ∈ R

so that f(c) is the global minimum value of f. Consider

f(x) = 4x^{2} – 4x + 11 = (2x – 1)^{2} + 10 ≥ ∀ x ∈ R ……………(1)

Now, f(1/2) = 10

Also f(x) ≥ f(1/2) ∀ x ∈ R

Hence, f(1/2) = 10 is the global minimum value of f(x), and a point of global minimum is x = 1/2.

Question 52.

Let f : [-2, 2] → R be defined by f(x) = |x|. Find the global maximum of f(x) and a point of global minimum.

Solution:

We know that |x| = \(\left\{\begin{array}{ccc}

x & \text { if } & x \geq 0 \\

-x & \text { if } & x<0

\end{array}\right.\)

Therefore, from the graph of the function f on [-2, 2] clearly f(x) ≤ f(2) and f(x) ≤ f(-2) ∀ x ∈ [-2, 2].

∴ f(2) = f(-2) = 2 is the global maximum of f(x), 2 and -2 are the points of global maximum.

Question 53.

Find the global maximum and global minimum of the function f : R → R defined by f (x) = x^{2}.

Solution:

We have f(x) ≥ f(0) ∀ x ∈ R.

Hence th global minimum value of f(x) is 0 and a point of global minimum is x = 0.

Suppose f has global maximum at x_{0} ∈ R (x_{0} > 0). Then as per out assumption we have.

f(x_{0}) ≥ f(x) ∀ x ∈ R

Choose x_{1} = x_{0} + 1. Then x_{1} ∈ R and x_{0} < x_{1}

∴ x_{0}^{2} < x_{1}^{2}

Hence f(x_{0}) < (fx_{1})

Thus we got f(x_{1}) such that f(x_{0}) > f(x_{0}) which is a contradition to Therefore, f(x) has no global maximum on R.

Question 54.

Find the stationary points of f(x) = 3x^{4} – 4x^{3} + 1, ∀ x ∈ R and state whether the function has local maxima or local minima at those points.

Solution:

Given that f(x) = 3x^{4} – 4x^{3} +1 and the domain of f is R. Differentiating the function w.r.t. x we have

f'(x) = 12x^{2}(x – 1) ……………………… (1)

The stationary points are the roots of f (x) = 0 i.e., 12x^{2}(x – 1) = 0. Hence x = 0 and x = 1 are the stationary points. Now, we test whether the stationary point x = 1 is a local extreme point or not. For.

f'(0.9) = 12(0.9)^{2} (0.9 – 1) ⇒ f'(0.9) is negative

f'(1.1) = 12(1.1)^{2} (1.1 – 1) ⇒ f'(1.1) is positive

and f(x) is defined in the neighbourhood i.e., (0.8, 1.2) of x = 1 with 8 = 0.2.

By theorem cis a point of local maximum if f(x) changes sign from positive to negative at x = c.

c is a point of local minimum if f'(x) changes sign from negative to positive at x = c.

The given function has local (relative) minimum at x = 1. Hence x = 1 is a local extreme point.

We will now test whether x = 0 is a local extreme point or not.

The function f(x) is defined in the neighbourhood of (-0.2, 0.2).

f'(-0.1) = 12(-0.1)^{2} (-0.1 – 1) ⇒ f'(-0.1) is negative

f'(-0.1) = 12(0.1)^{2} (0.1 -1) ⇒ f'(0.1) is negative

Thus, f(x) has no change in sign at x = 0. Therefore, the function f has no local maximum and no local minimum. Hence, x = 0 is not a local extreme point.

Question 55.

Find the points (if any) of local maxima and local minima of the function f (x) = x^{3} – 6x^{2} + 12x – 8 ∀ x ∈ R.

Solution:

Given function is f(x) = x^{3} – 6x^{2} + 12x – 8 and the domain of f is R.

Differentiating the given function w.r.t. x, we get

f'(x) = 3x^{2} – 12x + 12 i.e., f'(x) = 3(x – 2)^{2}.

The stationary point of f(x) is x = 2, since 2 is a root of f'(x) = 0.

Choose δ = 0.2 The 0.2- neighbourhood of 2 is (1.8, 2.2). Now

f'(1.9) = 3(1.9 – 2)^{2} ⇒ f'(1.9) is positive

f'(2.1) = 3(2.1- 2)^{2} ⇒ f'(2.1) is positive

Thus f(x) does not change the sign at x = 2. By Theorem c is neither a point of local maximum nor a point of local minimum if f'(x) does not change sign at x = c.

x = 2 is neithere a local maximum nor a local minimum.

Question 56.

Find the points of local minimum and local maximum of the function f(x) = sin 2x ∀ x ∈ [0, 2π]

Solution:

The given function is f(x) = sin 2x and domain is [0, 2π].

f'(x) = 2cos 2x …………… (1)

The critical points are the roots of 2 cos 2x = 0 and lying in the domain [0, 2π].

They are \(\frac{\pi}{4}\) and \(\frac{3\pi}{4}\) .

Now we apply the first derivative test at x = \(\frac{\pi}{4}\)

Clearly (\(\frac{\pi}{4}\) – 0.1 . \(\frac{\pi}{4}\) + 0.1) is a neighbourhood of \(\frac{\pi}{4}\) and the given f is defined on it.

Now

f'(\(\frac{\pi}{4}\) – 0.05) = 2 cos(\(\frac{\pi}{2}\) – 0.1) > 0

f'(\(\frac{\pi}{4}\) + 0.05) = 2 cos(\(\frac{\pi}{2}\) + 0.1) < 0

Thus f'(x) changes sign from positive to negative at x = \(\frac{\pi}{4}\). Therefore f has a local maximum.

Now we apply the first derivative test at x = \(\frac{3\pi}{4}\).

Clearly (\(\frac{3\pi}{4}\) – 0.1 . \(\frac{3\pi}{4}\) + 0.1) is a neighbourhood of \(\frac{3\pi}{4}\) and the given f is defined on it.

Now

f'(\(\frac{3\pi}{4}\) – 0.05) = 2 cos(\(\frac{3\pi}{4}\) – 0.1) < 0 f'(\(\frac{3\pi}{4}\) + 0.05) = 2 cos(\(\frac{3\pi}{4}\) + 0.1) > 0

Thus f'(x) changes sign from positive to negative at x = \(\frac{3\pi}{4}\). Therefore f has a local maximum at x = \(\frac{3\pi}{4}\).

Question 57.

Find the points of local extrema of the function f(x) = x^{3} – 9x^{2} – 48x + 6 ∀ x ∈ R Also find its local extrema.

Solution:

Given function is

f(x) = x^{3} – 9x^{2} – 48x + 6 …………… (1)

and the domain of the function is R.

Differentiating (1) w.r.t. x we get

f'(x) = 3x^{2} – 18x – 48 = 3(x – 8) (x + 2) ………….. (2)

Thus the stationary points are – 2 and 8.

Differentiating (2) w.r.t.x we get,

f'(x) = 6(x – 3) ………….. (3)

Let x_{1} = -2 and x_{2} = 8. Now we have to find f’ at each of these points to know the sign of second derivative.

At x_{1} = -2, f'(-2) = – 30. The sign of it is negative.

∴ x_{1} = – 2 is a point of local maximum of f and its local maximum value is f(-2) – 58.

Now, at x_{2} = 8 f'(8) = 30. Thus the sign of f”(x_{2}) is positive. Therefore, x = 8 is a point of local minimum of f and its local minimum value is

f(8) = – 442.

Question 58.

Find the points of local extrema of f(x) = x^{6} ∀ x ∈ R. Also find its local extrema.

Solution:

f(x) = x^{6} ………………….. (1)

Differentiating (1) w.r.t. x we get,

f'(x) = 6x^{5} ……………. (2)

and again differentiating (2) w.r.t. x we get

f'(x) = 30x^{4} …………………… (3)

The stationary point of f(x) is x = 0 only (since f'(x) = 0 only at x = 0).

Now f'(0) = 0. At x = 0, we can not conclude anything about the local extrema by the second derivative test. Therefore, we apply the first derivative test. As the domain of f is R, the function f is defined on (-0.2, 0.2) which is a neighbourhood of x = 0. Now

f'(-0.1) = 6(-0.1)5 < 0, f'(0, 1) = 6(0.1) 5 > 0.

Thus f'(x) changes sign form negative of positive at x = 0.

∴ x = 0 is a point of local minimum and its local minimum value is f(0) = 0.

Question 59.

Find the points of local extrema and local extrema for the function f(x) = cos 4x defined on (0, \(\frac{\pi}{2}\))

Solution:

Here f(x) = cos 4x ……………. (1)

and its domain is (0, \(\frac{\pi}{2}\))

∴ f’(x) = -4 sin 4x ………….. (2)

and f”(x) = -16 cos 4x ……………… (3)

The stationary points are the roots of

f'(x) = 0 and lying in the domain (o, \(\frac{\pi}{2}\)).

f'(x) = 0 ⇒ 4 sin 4x = 0

⇒ 4x = 0, π, 2π, 3π, 4π ………………….

⇒ x = 0, π/4, π/2, 3π/4, π …………………..

The point lying in the domain is x = \(\frac{\pi}{4}\) only.

Thus x = \(\frac{\pi}{4}\) is the stationary point of the given function. Now

f”(\(\frac{\pi}{4}\)) = -16 cos(π) = 16 > 0.

The function f has local minimum at x = \(\frac{\pi}{4}\) and its local minimum value is

f(\(\frac{\pi}{4}\)) = -1.

Question 60.

Find two positive number whose sum is 15 so that the sum of their squares is minimum.

Solution:

Suppose one numbers is x and the other number 15 – x. Let S be the sum of squares of these numbers. Then S = x^{2} + (15 – x)^{2} ………………… (1)

Note that the quantity S, to be minimized, is a function of x.

Differentiating (1) w.r.t. x, we get

\(\frac{\mathrm{dS}}{\mathrm{dx}}\) = 2x + 2(15 – x) (-1)

= 4x – 30 ………………. (2)

and again differentiating (2) w.r.t.x, we get

\(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\) = 4 ……………….. (3)

The stationary point can be obtained by solving \(\frac{\mathrm{dS}}{\mathrm{dx}}\) = 0 i.e., 4x – 30 = 0.

∴ x = 15/2 is the stationary point of (1).

Since \(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\) = 4 > 0, S is minimum at x = \(\frac{15}{2}\)

∴ The two numbers are \(\frac{15}{2}\), 15 – \(\frac{15}{2}\) i.e., \(\frac{15}{2}\) and \(\frac{15}{2}\).

Question 61.

Find the maximum area of the rectangle that can be formed with fixed perimeter 20.

Solution:

Let x and y denote the length and the breadth of a rectangle respectively. Given that the perimeter of the rectangle is 20.

i.e., 2(x + y) = 20

i.e., x + y = 10 ……………. (1)

Let A denote the area of rectangle.

Then A = xy ………….. (2)

Which is to be minimized. Equation (1) can be expressed as

y = 10 – x …………… (3)

From (3) and (2), we have

A = x (10 – x)

A = 10x – x^{2} ……………… (4)

Differentiating (4) w.r.t. x we get

\(\frac{\mathrm{dA}}{\mathrm{dx}}\) = 10 – 2x ……………….. (5)

The stationary point is a root of 10 – 2x = 0

∴ x = 5 is the stationary point.

Differentiating (5) w.r.t. x, we get

\(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\) = -2

which is negative. Therefore by second derivative test the area A is maximized at x = 5 and hence y = 10 – 5 = 5, and the maximum area is A = 5(5) = 25.

Question 62.

Find the point on the graph y^{2} = x which is the nearest to the point (4, 0).

Solution:

Let P(x, y) be any point on y^{2} = x and A(4, 0). We have to find P such that PA is minimum

Suppose PA = D. The quantity to be minimized is D.

D = (\(\sqrt{(x-4)^{2}+(y-0)^{2}}\)) ………………. (1)

P(x, y) lies on the curve, therefore

y^{2} = x …………………… (2)

From (1) and (2), we have

D = \(\sqrt{\left((x-4)^{2}+x\right)}\)

D = \(\sqrt{\left(x^{2}-7 x+16\right)}\) ……………………. (3)

Differentiating (3) w.r.t. x, we get

\(\frac{\mathrm{dD}}{\mathrm{dx}}\) = \(=\frac{2 x-7}{2} \cdot \frac{1}{\sqrt{x^{2}-7 x+16}}\)

Now \(\frac{\mathrm{dD}}{\mathrm{dx}}\) = 0

gives x = \(\frac{7}{2}\). Thus \(\frac{7}{2}\) is a stationary point of the function D. We apply the first derivative test to verify whether D is minimum at x = \(\frac{1}{2}\)

Question 63.

Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

Solution:

Let O be the centre of the circular base of the cone and its height be h. Let r be the radius of the circular base of the cone.

Then AO = h, OC = r.

Let a cylinder with radius x(OE) be inscribed in the given cone. Let its height be u.

i.e., RO = QE = PD = u

Now the triangles AOC and QEC are similar.

Therefore,

\(\frac{\mathrm{QE}}{\mathrm{OA}}\) = \(\frac{\mathrm{EC}}{\mathrm{OC}}\)

i.e., \(\frac{u}{h}\) = \(\frac{r-x}{r}\)

∴ u = \(\frac{h(r-x)}{r}\) ………………… (1)

Let S denote the curved surface area of the chosen cylinder.Then

S = 2 π xu.

As the cone is fixed one, the values of r and h are constants. Thus S is function of x only. Now,

\(\frac{\mathrm{dS}}{\mathrm{dx}}\) = 2 πh (r – 2x)/r and \(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\) = -4πh/r.

The stationary point of S is a root of

\(\frac{\mathrm{dS}}{\mathrm{dx}}\) = 0

i.e., π(r – 2x)/r = 0

i.e., x = r/2

\(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\) < 0 for all x, Therefore (\(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\))_{x=r/2} < 0.

Hence, the radius of the cylinder of greatest curved surface area which can be inscribed in a given cone is r/2.

Question 64.

The profit function P(x) of a company, selling x items per day is given by P(x) = (150 – x)x – 1600. Find the number of items that the company should sell to get maximum profit. Also find the maximum profit.

Solution:

Given that the profit function is

P(x) = ( 150 – x)x – 1600 ………….. (1)

For maxima or minima \(\frac{\mathrm{dP}(\mathrm{x})}{\mathrm{dx}}\) = 0

∴ (150 – x) (1) + x (-1) = 0

i.e., x = 75

Now \(\frac{d^{2} P(x)}{d x^{2}}\) = -2 and \(\left[\frac{d^{2} P(x)}{d x^{2}}\right]_{x=75}\) < 0.

∴ The profit P(x) is maximum for x = 75.

∴ The company should sell 75 items a day to make maximum profit.

The maximum profit will be P(75) = 4025.

Question 65.

A manufacturer can sell x items at a price of rupees (5 – x/100) each. The cost price of x items is Rs. (x/5 + 500). Find the number of items that the manufacturer should sell to earn maximum profits.

Solution:

Let S(x) be the selling price of x items and C(x) be the cost price of x items. Then, we have

S(x) = {cost of each item}, x .

∴ S(x) = (5 – x/100) x = 5x – x^{2}/100 and C(x) = x/5 + 500

Let P(x) denote the profit function. Then

P(x) = S(x) – C(x)

P(x) = (5x – x^{2}/100) – (x/5 + 500)

– (24x/5) – (x^{2}/100) – 500 …………….. (1)

For maxima or minima

\(\frac{\mathrm{dP}(\mathrm{x})}{\mathrm{d} x}\) = 0

i.e., 24/5 – x/50 = 0

The stationary point of P(x) is x = 240 and

\(\left[\frac{d^{2} P(x)}{d x^{2}}\right]\) = –\(\frac{1}{50}\) for all x.

Hence the manufacturer can earn maximum profit if he sells 240 items.

Question 66.

Find the absolute extrement of f(x) = x^{2} defined on [-2, 2].

Solution:

The given function f(x) = x^{2} is continuous on [-2, 2]. It can be shown that it has only local minimum and the point of local minimum is 0. The absolute(global) maximum of f is the largest value of f(-2), f(0) and f(2) i.e., 4, 0, 4.

Hence, the absolute maximum value is 4. Similarly the absolute minimum is the least value of 4, 0, 4. Hence 0 is the absolute minimum value.

Question 67.

Find the absolute maximum of x^{40} – x^{20} on the interval [0, 1]. Find also its absolute maximum value.

Solution:

Let f(x) = x^{40} – x^{20} ∀ x ∈ [0,1] ………… (1)

The function f is continuous on [0,1 j and the interval [0,1] is closed.

From (1) we have

f'(x) = 40 x^{39} – 20 x^{19} = 20x^{19} (2x^{20} – 1).

Thus f'(x) = 0 at x = 0 or

x = \(\left(\frac{1}{2}\right)^{\frac{1}{20}}\)

Therefore, the critical points of f are and \(\left(\frac{1}{2}\right)^{\frac{1}{20}}\) and 0 is one of the end points of the domain. Therefore no local maximum exists at x = 0. Now

f'(x) = 40(39) x^{38} – 20(19) x^{18}

= 20x^{18} (78 x^{20} – 19)

[f”(x)]_{x = \(\left(\frac{1}{2}\right)^{\frac{1}{20}}\)} = 20 (1/2)^{(18/20)}[39 – 19] > 0.

Therefore f has local minimum at

x = (1/2)(1/20)

and its value is \(f\left(\left(\frac{1}{2}\right)^{\frac{1}{20}}\right)\) = –\(\frac{1}{4}\)

Therefore the absolute maximum value of the function f is the largest value of f(0), f(1) and \(f\left(\left(\frac{1}{2}\right)^{\frac{1}{20}}\right)\) i.e., the largest value of {0, 0, –\(\frac{1}{4}\)}

Hence, the absolute maximum of f is 0 and the points of absolute maximum are 0 and 1. Further the absolute minimum is the least of 0, 0, -latex]\frac{1}{4}[/latex].

Hence the absolute minimum is -latex]\frac{1}{4}[/latex] and the point of absolute minimum is x = \(\left(\frac{1}{2}\right)^{\frac{1}{20}}\)